MIT OpenCourseWare
http://ocw.mit.edu
5.111 Principles of Chemical Science
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
5.111 Lecture Summary #12
Readings for today: Section 2.9 (2.10 in 3rd ed) , Section 2.10 (2.11 in 3rd ed), Section
2.11 (2.12 in 3rd ed), Section 2.3 (2.1 in 3rd ed), Section 2.12 (2.13 in 3rd ed).
Read for Lecture #13: Section 3.1 (3rd or 4th ed) – The Basic VSEPR Model, Section 3.2
(3rd or 4th ed) – Molecules with Lone Pairs on the Central Atom.
_______________________________________________________________________________
Topics:
I. Breakdown of the octet rule
Case 1. Odd number of valence electrons
Case 2. Octet deficient molecules
Case 3. Valence shell expansion
II. Ionic bonds
III. Polar covalent bonds and polar molecules
________________________________________________________________________________
I. BREAKDOWN OF THE OCTET RULE
Case 1. Odd number of valence electrons
For molecules with an odd number of valence electrons, it is not possible for each
atom in the molecule to have an octet, since the octet rule works by ____________ e-s.
Example: CH3
H
2)
3(1) + 4 = ______ valence electrons
3)
3(2) + 8 = ______ electrons needed for octet
4)
14 – 7 = ________ bonding electrons
H3 C
H C H
Radical species: molecule with an ________________ electron.
Radicals are usually very reactive. The reactivity of radical species leads to
interesting (and sometimes harmful) biological activity.
Free radicals in biology: a paradox
Free radical species damage
DNA. in biology: a Free
radicals are essential for life.
Free radicals
paradox
Free radicalivespecies
DNA.
oxygendamage
radicals are
a byproduct
• Highly reactive oxygen radicals are a byproduct
of metabolism and cause DNA damage.
• Cigarette smoke contains free radicals that can
damage DNA in lung cells and leadantioxidant-rich
to cancer.
antioxidant-rich
lueberries
oxidants (such as vitaminbA,
C, and E) reduce
• Antioxidants (such as vitamin A, C, and E) reduce
DNA damage in the body by “trapping” radicals.
Free radicalsFree
areradicals
essential
for life.in cell
are involved
• Free radicals are involved in cell
(see NO use
example
on p. 2).to
signaling
Some enzymes
free radicals
carry out essential reactions in the
enzymes use free radicals to
• Some
body.
carry out essential reactions in the
dy. called ribonucleotide reductase
bootein
For example, a pr
(RNR) catalyzes an essential step in DNA synthesis
Forand
example,
protein
called
ribonucleotide
reductase
repair ausing
a free
radical
species.
(RNR) catalyzes an essential step in DNA synthesis
and repair using a free radical species.
Some radicals are more stable. For example, NO
NO
N
O
1)
Draw skeletal structure
2)
5 + 6 = 11 valence electrons
3)
8 + 8 = 16 electrons needed for octet
4)
16 – 11 = ______ bonding electrons
Nitric oxide, NO
NO
vasodilation:
the widening of blood vessels
• an important cell-signaling molecule in humans.
• diffuses freely across cell membranes and signals for the smooth muscle in
blood vessels to relax, resulting in vasodilation and increased blood flow.
• a radical species, NO has a short lifetime in the body, which makes it an ideal
messenger molecule between adjacent cells.
• You may be familiar with a drug that inhibits the breakdown of an NO binding
partner (an enzyme), leading to increased blood flow: _________________.
Figure by MIT OpenCourseWare.
Now let’s think about molecular oxygen, O2.
What we expect: O O
2)
_______________ valence electrons
3)
_______________ electrons needed for octet
4)
_______________ bonding electrons
5)
Add two electrons per bond.
6)
2 bonding electrons remaining. Make double bond.
7)
_______________ valence electrons – make lone pairs
Lewis method seems to work here, but in reality O2 is a ________________!
O O
We need molecular orbital (MO) theory (Lecture #14).
2
Case 2. Octet deficient molecules
Some molecules are stable with an incomplete octet. Group 13 elements ____ and
_____ have this property.
Consider BF3
F
F B F
First, let’s write the Lewis structure that achieves octets on every atom. 2)
3 + 3(7) = ______ valence electrons 3)
8 + 3(8) = ______ electrons needed for octet 4)
32 – 24 = ________ bonding e-s
5)
assign two electrons per bond. 6)
8 – 6 = 2 extra bonding electrons
7)
24 – 8 = 16 lone pair electrons 8)
calculate formal charges: FCB = 3 – 0 – (½)(8) = -1
FCFDB = 7 – ____ – (½)(____) = _______
FCF = 7 – ____ – (½)(____) = _______
But experiments suggest that all three B-F bonds have the same length, that of a
______________ bond.
F
F B F
FCB = 3 – 0 – (½)(6) = 0
FCF = 7 – 6 – (½)(2) = 0
The formal charges are more favorable for this structure.
Case 3. Valence shell expansion
Elements with n = or > 3 have empty ____ - orbitals, which means more than eight
electrons can fit around the central atom.
Expanded valence shells are more common when the central atom is ______ and is
bonded to small, highly electronegative atoms such as O, F, and Cl.
3
Consider PCl5
Cl
Cl P
Cl
Cl Cl
2)
5 + 5(7) = ______ valence electrons
3)
8 + 5(8) = ______ electrons needed for octet
4)
48 – 40 = ______ bonding e-s
To make five P-Cl bonds, need ______ shared electrons. So 40 – 10 = 30 lone-pair
electrons.
Consider CrO42­
-2
O
O
2)
6 + 4(6) + _____ = ______ valence electrons
Cr O
3)
8 + 4(8) = ______ electrons needed for octet
O
8)
4)
40 – 32 = ______ bonding e-s
7)
32 – 8 = 24 lone-pair electrons.
calculate formal charges:
FCCr = 6 – 0 – (½)(8) = +2
FCO = 6 – 6 – (½)(2) = -1
Total charge = 2 + 4(-1) = -2
But experimentally, Cr-O bond length and strength are between that of a single and
double bond!
-2
-2
O
O
Cr O
O
O
O
Cr O
O
plus ________ other resonance
structures.
FCCr = 6 – 0 – (½)12 = 0
FCODB = 6 – 4 – (½)4 = 0
FCO = 6 – 6 – (½)2 = -1
4
Valence shell expansion around Cr results in ________ formal charge separation.
More stable Lewis structure.
II. IONIC BONDS
Ionic bonds involve the complete _____________ of (one or more) electrons from one
atom to another with a bond resulting from the electrostatic attraction between the
cation and anion.
Consider the formation of NaCl from the neutral atoms, Na and Cl.
Na (g)
Na+ (g) + e–
∆E = ______ = 494 kJ/mol
Cl + e–
Cl– (g)
∆E = – EA = _____ kJ/mol
Na+ (g) + Cl– (g)
∆E = + ________ kJ/mol
Na (g) + Cl (g)
∆E = IENa + (– EACl) = 145 kJ/mol
Problem: Na (g) + Cl (g) ⇒ Na+ (g) + Cl– (g) has a positive ∆E. It ___________ energy.
Solution: Coulomb attraction.
Na+ (g) + Cl– (g)
NaCl (g)
∆E = – 589 kJ/mol
Net change in energy for the overall process:
Na (g) + Cl (g)
NaCl (g)
∆E = __________ kJ/mol
The mutual attraction between the oppositely-charged ions releases energy. The net
energy change for the formation of NaCl is a decrease in energy.
We can calculate the Coulomb attraction based on the distance between the two ions
(assume here that the ions are point charges):
U(r) =
z 1z 2e 2
4πε0r
for 2 unlike charges,
z = charge numbers of the ions and
e = absolute value of the charge of an e- (1.602 X 10-19 C)
Calculate U(r) for Na+ and Cl-. NaCl has a bond length (r) = 2.36Å.
5
U(r) = (
)(
)(
4π(8.854 x 10-12 C2J-1m-1)(
) =
)
Convert to kJ/mol
U(r) = -9.774 x 10-19 J x _________ x _______________ =
Simple ionic model predicts:
∆E = -∆Ed = - 444 kJ/mol
Experiments measure:
∆E = -∆Ed = - 411 kJ/mol
The discrepancy results from the following approximations:
• ignored repulsive interactions. Result:_________ ∆Ed than experimental value.
• treated Na+ and Cl– as ________________________.
• ignored quantum mechanics.
This simple model is applicable only to very ionic bonds.
III. POLAR COVALENT BONDS
Perfectly-ionic and perfectly-covalent bonds are the two extremes of bonding. In
reality, most bonds fall somewhere in the middle.
A polar covalent bond is an ____________ sharing of electrons between two atoms
with different electronegativities (χ).
Consider H-Cl versus H-H (Pauling electronegativity values are given):
χH = 2.2
H – Cl
δ
χCl = 3.2
δ–
H +– Cl
where δ is fraction of a full charge (e) that is asymmetrically distributed.
H–H
H2 is a “perfectly” covalent bond, δ = 0.
6
Dipole moment
Asymmetric charge distribution results in an electric dipole, two unlike charges
separated by a finite distance.
We can quantify charge separation by defining a dipole moment, µ .
C m
δ−
δ+
Dipole moment is measured in C•m or in debye
μ→= Q•→
r
r
Magnitude of charge
Separation where Q =⏐δe⏐
→
μ
1 debye = 1 D = 3.336 x 10–30 C•m In chemistry, the arrow points toward the negative charge in a polar bond.
Polar molecules have a non-zero net dipole moment.
χH = 2.2
χC = 2.6
O
C
O
χN = 3.0
O
H
H
H2O
CO2
χO = 3.4
molecule
molecule
In large organic molecules and in biomolecules, such as proteins, we often consider the number of polar groups within the molecule. For example, let’s compare vitamin A to vitamin B9
Which vitamin contains a higher number of polar bonds? vitamin ________
HO
HO
H
C
H33C
CH
CH33
CH
CH33
H
H
C
C
H
H22C
C
C
C
H
H22C
C
C
C
C
C
H
H22
H
H
C
C
C
C
H
H
O
O
C
C
CH
CH33
C
C
H
H
H
H22
C
C
C
C
C
C
H
H
C
C
H
H
HC
HC
H
H
H
C
H22C
O
O
O
O
CH
CH22
C
C
CH
CH33
O
O
OH
OH
H
H
C
C
C
C
N
N
H
H
C
C
H
H
CH
CH
HC
HC
O
O
H
H22
C
C
C
C
C
C
H
H
N
N
H
H
N
N
C
C
C
C
C
C
HC
HC
C
C
N
N
Vitamin A
________________ soluble
N
N
C
C
NH
NH22
N
N
Vitamin B9 (_____________________)
________________ soluble
7